Two satellites $S_{1}and{S}_2$ revolve around a planet in coplanar circular orbits in the same sense. Their periods of revolution are 1 hour and 8 hours respectively. The radius of orbit of  $S_1 is {10}^{4}km$. When $S_2$ is closest to $S_1$, The angular speed of $S_2$ as actually observed by an astronaut in $S_{1}is \frac{\pi }{X}$ rad/hr. What is the value of X?

As according to Kepler’s 3rd law of planetary motion, so we have
$$\begin{gathered} {\left[\frac{T_1}{T_2}\right]}^2={\left[\frac{r_1}{r_2}\right]}^3 \\ i.e.,{ \left[\frac{1}{8}\right]}^2={\left[\frac{{10}^4}{r_2}\right]}^3 \\ or, r_2=4\times {10}^{4}km \end{gathered}$$
Time period of a body in circular motion is
$$\begin{gathered} T=\frac{2\pi r}{v},V=\frac{2\pi r}{T} \\ So, |{\overline{v}}_2-{\overline{v}}_1|=2\pi [\frac{r_1}{T_1}-\frac{r_2}{T_2}] \\ i.e.,|{\overline{v}}_2-{\overline{v}}_1|=2\pi |\frac{4\times {10}^4}{8}~\frac{{10}^4}{1}|=\pi \times {10}^{4}km/hr \end{gathered}$$

b) When they are closest to each other and moving in the same direction.

$$\begin{gathered} \left|v_{net}\right|=|{\overline{v}}_2-{\overline{v}}_1|=\pi \times {10}^{4}km/hr \\ and{\overline{r}}_{net}=|{\overline{r}}_2-{\overline{r}}_1|=(4\times {10}^4-1\times {10}^4) \\ =3\times {10}^{4}km \\ So, {\omega }_{net}=\frac{\left|v_{net}\right|}{\left|{\overline{r}}_{net}\right|}=\frac{\pi \times {10}^4}{3\times {10}^4}=\frac{\pi }{3}rad/hr \end{gathered}$$